Chapter 6.1, Problem 11WE

To determine

To calculate: The simplified form of the expression $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}$ and state restrictions on variable.

Answer to Problem 11WE

The simplified value of the expression $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}$ is $\frac{x+4}{x-3}$ and restriction is $x\ne 3,x\ne -\frac{1}{2}$.

Explanation of Solution

Given information:

The expression $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}$.

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

Consider the expression $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}$.

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, strike out common terms from the given expression and simplify it as,

  $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}=\frac{x+4}{x-3}$

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\left(1+2x\right)\left(x-3\right)=0$

There are two possibilities either $\begin{array}{c}1+2x=0\\ 2x=-1\\ x=-\frac{1}{2}\end{array}$ or $\begin{array}{c}x-3=0\\ x=3\end{array}$.

For denominator not to be zero, $x\ne 3,x\ne -\frac{1}{2}$.

Thus, the simplified value of the expression $\frac{\left(x+4\right)\left(2x+1\right)}{\left(1+2x\right)\left(x-3\right)}$ is $\frac{x+4}{x-3}$ and restriction is $x\ne 3,x\ne -\frac{1}{2}$.